Equilibria.... Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion:

Question

Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion:

Fe3 (aq) 3 C2O42-(aq) [Fe(C2O4)3]3-(aq).

If 0.0100 M Fe3 is initially mixed with 1.00 mol dm-3 oxalate ion, what is the concentration of Fe3 ion at equilibrium?

a. 1.67 × 1018 mol dm-3

b. 6.56 × 10-23mol dm-3

c. 0.0100 mol dm-3

d. 1.62 × 1022 mol dm-3

Answer 1

Fe3+   &   3 (C2O4)2-    ---> [Fe(C2O4)3]3-

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with a K like 1.67e20, we assume that essentially all of the 0.01 Fe+3 became [Fe(C2O4)3]3-, that whatever  Fe+3 remains in the solution at equilibrium is "X",

and that oxalate which was once 1.00 has decreased by 3 times the 0.01moles that went w/ the Fe) to 0.97,  (1:3 ratio combine)

and finally that whatever variance from 0.97 Molar C2O4 is insignificant (so I won't use the 0.97 +3X, that it really is)

K =  [Fe(C2O4)3] / [Fe3+] [C2O4]^3

1.67 e20  =  [0.0100] / [X] [0.97]^3

X =   [0.0100] / [0.97]^3 (1.67 e20)

X = 0.0100 / 1.524e20

X = 6.56 e-23 Molar Fe+3

that's your answer: 6.56 e-23 Molar Fe+3